Answer to Question #152991 in Mechanics | Relativity for wisdom kutubu

Question

The position of a particle moving along x axis is given by x=12t^2-2t^3, where x is in meters and t is in seconds. Determine (a) The position, (b) The velocity, (c) the acceleration of the particle at t=3.0 s. (d) what is the maximum positive coordinate reached by the particle (e) at what time did it reached ? (f) what is the maximum positive velocity reached by the particle (g) at what time did it reached? (h) what is the acceleration of the particle at the instant the particle is not moving other than at t=3 s.

Answer 1

Solution.

(a) the position of the particle $x=12t^2-2t^3$

when time (t)=3s $x(3)=12(3^2)-2(3^3)$

$=54m$

(b) the velocity of the particle is given by the derivative of x.

therefore when $x=12t^2-2t^3$

$v=dx/dt$

$v=24t-6t^2$

$v(3)=24*3-6(3^2)$

$v=72-54$

$v=18m/s$

(c) the acceleration of the particle is given by the derivative of the velocity

$a=dv/dt$

$a=24-12t$

$a(3)=24-12(3)$

$a=-12m/s^2$

(d) the maximum positive coordinate reached by the particle.

this is when $dx/dt=0$

hence, $dx/dt=24t-6t^2=0$

by factorizing,

$=6t(4-t)=0$

so therefore, either

$t=0$$s$

or

$t=4s$

in ths case therefore the maximum coordinate is given when $x=4$

$x=12(4^2)-2(4^3)$

$x=64m$

(e) the maximum value reached when $t=4s$ we have found this from part (d) above.

(f) the maximum positive velocity reached by the particle is when the derivative of $v(t)$ is equal to zero.

$dv/dt=24-12t=0$

hence,

$t=2s$

therefore $v(t)$ reaches maximum when $t=2s$

so, $v=24(2)-6(2^2)=24m/s$

(g) from part (f) we have seen that velocity the maximum value when time $($ $t)=2s$

(h) the acceleration of the particle at the instant the particle is not moving other than at t=3 s.

when $dx/dt(t)=0$ at $t=4s$ derived from part (d)

hence, acceleration at this time is given by

$a=24-12(4)=-24m/s^2$

in this case, when $t=0$ the particle is at position $x=0$ and $t=3$ hence the particleis at position $x=54m$

therefore the average velocity is given by $change (x)/change(t)$ i.e $\Delta$ $x/\Delta$ $t$

V_avg$=54-0$ $/3-0$

$=54/3$

$=18m/s$